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Using the provided Venn diagram and definitions, identify equivalent sets and calculate the square root of the given numbers using prime factorization.

Mathematics

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6 steps
  1. Solve Question 1.

    • Definition of a proper subset: A set AA is a proper subset of a set BB if all elements of AA are also elements of BB, and BB contains at least one element not in AA. This is denoted as A⊂BA \subset B.
    • Elements of sets A, B, and C from the diagram:
      • A={b,c,d,e,f}A = \{b, c, d, e, f\}
      • B={d,e,f,g}B = \{d, e, f, g\}
      • C={f,g,h,i}C = \{f, g, h, i\}
    • Identify equivalent sets: Two sets are equivalent if they have the same number of elements (cardinality).
      • Number of elements in AA, ∣A∣=5|A| = 5.
      • Number of elements in BB, ∣B∣=4|B| = 4.
      • Number of elements in CC, ∣C∣=4|C| = 4.
      • Since ∣B∣=∣C∣|B| = |C|, sets BB and CC are equivalent sets.
      • The equivalent sets are B and C\boxed{B \text{ and } C}.
  2. Solve Question 2 by finding the square root using prime factorization.

    • a) For 50176: 50176=2×2508850176 = 2 \times 25088 =2×2×12544= 2 \times 2 \times 12544 =2×2×2×6272= 2 \times 2 \times 2 \times 6272 =2×2×2×2×3136= 2 \times 2 \times 2 \times 2 \times 3136 =2×2×2×2×2×1568= 2 \times 2 \times 2 \times 2 \times 2 \times 1568 =2×2×2×2×2×2×784= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 784 =2×2×2×2×2×2×2×392= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 392 =2×2×2×2×2×2×2×2×196= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 196 =2×2×2×2×2×2×2×2×2×98= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 98 =2×2×2×2×2×2×2×2×2×2×49= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 49 =210×72= 2^{10} \times 7^2 To find the square root, we take half of each exponent: 50176=210×72=210/2×72/2=25×71\sqrt{50176} = \sqrt{2^{10} \times 7^2} = 2^{10/2} \times 7^{2/2} = 2^5 \times 7^1 =32×7=224= 32 \times 7 = 224 The square root of 50176 is 224\boxed{224}.
    • b) For 105625: 105625=5×21125105625 = 5 \times 21125 =5×5×4225= 5 \times 5 \times 4225 =5×5×5×845= 5 \times 5 \times 5 \times 845 =5×5×5×5×169= 5 \times 5 \times 5 \times 5 \times 169 =54×132= 5^4 \times 13^2 To find the square root, we take half of each exponent: 105625=54×132=54/2×132/2=52×131\sqrt{105625} = \sqrt{5^4 \times 13^2} = 5^{4/2} \times 13^{2/2} = 5^2 \times 13^1 =25×13=325= 25 \times 13 = 325 The square root of 105625 is 325\boxed{325}.
  3. Solve Question 3 by finding the cube number of the given numbers.

    • a) For 22: The cube number of 22 is 22322^3. 223=22×22×2222^3 = 22 \times 22 \times 22 =484×22= 484 \times 22 =10648= 10648 The cube number of 22 is 10648\boxed{10648}.
    • b) For 44: The cube number of 44 is 44344^3. 443=44×44×4444^3 = 44 \times 44 \times 44 =1936×44= 1936 \times 44 =85184= 85184 The cube number of 44 is 85184\boxed{85184}.
  4. Solve Question 4 by dividing Rs.

    100 between Ram and Hari in the ratio 2:3.

    • Total amount = Rs. 100
    • Ratio of shares for Ram and Hari = 2:3
    • Sum of the ratio parts = 2+3=52 + 3 = 5
    • Ram's share: Ram’s share=25×100=2×20=40\text{Ram's share} = \frac{2}{5} \times 100 = 2 \times 20 = 40
    • Hari's share: Hari’s share=35×100=3×20=60\text{Hari's share} = \frac{3}{5} \times 100 = 3 \times 20 = 60
    • Ram gets Rs. 40\boxed{\text{Rs. } 40} and Hari gets Rs. 60\boxed{\text{Rs. } 60}.
  5. Solve Question 5 by simplifying the expressions.

    • i) (3m)4×(3m)−3×(3m)−5(3m)−2×(3m)−3\frac{(3m)^4 \times (3m)^{-3} \times (3m)^{-5}}{(3m)^{-2} \times (3m)^{-3}} Using the exponent rule ax×ay=ax+ya^x \times a^y = a^{x+y}: =(3m)4+(−3)+(−5)(3m)(−2)+(−3)= \frac{(3m)^{4 + (-3) + (-5)}}{(3m)^{(-2) + (-3)}} =(3m)4−3−5(3m)−2−3= \frac{(3m)^{4 - 3 - 5}}{(3m)^{-2 - 3}} =(3m)−4(3m)−5= \frac{(3m)^{-4}}{(3m)^{-5}} Using the exponent rule axay=ax−y\frac{a^x}{a^y} = a^{x-y}: =(3m)−4−(−5)= (3m)^{-4 - (-5)} =(3m)−4+5= (3m)^{-4 + 5} =(3m)1= (3m)^1 =3m= 3m The simplified expression is 3m\boxed{3m}.
    • ii) aa−b×ab−c×ac−aa^{a-b} \times a^{b-c} \times a^{c-a} Using the exponent rule ax×ay×az=ax+y+za^x \times a^y \times a^z = a^{x+y+z}: =a(a−b)+(b−c)+(c−a)= a^{(a-b) + (b-c) + (c-a)} =aa−b+b−c+c−a= a^{a-b+b-c+c-a} =a0= a^0 Using the exponent rule a0=1a^0 = 1: =1= 1 The simplified expression is 1\boxed{1}.
  6. Solve Question 6 by simplifying the expression.

    • $\frac{1}{4} + \left
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